HDU 1711 Number Sequence （KMP）

题目链接

Problem Description
Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input
The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

Output
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

Sample Input
2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output
6
-1

分析 ：

kmp算法：
1 kmp是用来匹配字符串，只能够匹配单一的字符串
2 kmp的算法的过程：
1：假设文本串的长度为n，模式串的长度为m；
2：先例用O(m)的时间去预处理next数组，next数组的意思指的是当前的字符串匹配失败后要转到的下一个状态；
3：利用o(n)的时间去完成匹配；

代码：

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

#define MAXN 1000010

int t , n , m;
int text[MAXN];/*文本串*/
int pattern[MAXN];/*模式串*/
int Next[MAXN];/*Next数组*/

/*O(m)的时间求Next数组*/
void getNext(){
    Next[0] = Next[1] = 0;
    for(int i = 1 ; i < m ; i++){
       int j = Next[i];
       while(j && pattern[i] != pattern[j])
          j = Next[j];
       Next[i+1] = pattern[i] == pattern[j] ? j+1 : 0;
    }
}

/*o(n)的时间进行匹配*/
void find(){
    int j = 0;/*初始化在模式串的第一个位置*/
    for(int i = 0 ; i < n ; i++){/*遍历整个文本串*/
       while(j && pattern[j] != text[i])/*顺着失配边走，直到可以匹配，最坏得到情况是j = 0*/
         j = Next[j];
       if(pattern[j] == text[i])/*如果匹配成功继续下一个位置*/
         j++;
       if(j == m){/*如果找到了直接输出*/
         printf("%d\n" , i-m+2);/*输出在文本串中第一个匹配的位置，不是下标*/
         return;
       }
    }
    printf("-1\n");
}

int main(){
   scanf("%d" , &t);
   while(t--){
      scanf("%d%d" , &n , &m);
      for(int i = 0 ; i < n ; i++)
         scanf("%d" , &text[i]);
      for(int i = 0 ; i < m ; i++)
         scanf("%d" , &pattern[i]);
      getNext();
      find();
   }
   return 0;
}